Exercises — Aufbau principle — order of filling (Madelung rule, n + l)

Read the number as $n$; convert the letter to $l$ using $s=0,p=1,d=2,f=3$; add.

Compute $K$ for each; smaller $K$ wins; equal $K$ → smaller $n$ wins.

• (a) 2p: $K=2+1=3$; 3s: $K=3+0=3$. Tie! Smaller $n$ → 2p first.
• (b) 4s: $K=4$; 3d: $K=5$. 4s first.
• (c) 5s: $K=5$; 4d: $K=6$. 5s first.

Fill in Madelung order until 15 electrons are placed, using capacities $s=2,p=6$: $1s^{2}2s^{2}2p^{6}3s^{2}3p^3$ Check the count: $2+2+6+2+3=15$. ✅ $3p$ stops at $3$ because $2+2+6+2=12$ are already used, leaving $15-12=3$ for $3p$.

Madelung order puts 4s before 3d ($K_{4s}=4<K_{3d}=5$): $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^2$ Count: $2+2+6+2+6+2+2=22$. ✅ The nearest noble gas below is Argon ($Z=18$: $1s^2\dots 3p^6$). Everything up to $3p^6$ is $[\text{Ar}]$: $\boxed{[\text{Ar}]4s^{2}3d^2}$ Cosmetically re-ordered by $n$: $[\text{Ar}]3d^{2}4s^2$.

Sweep the staircase (Figure above) by increasing $K$, breaking ties by smaller $n$: $1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p$

Compute $K$: 3d $=3+2=5$, 4p $=4+1=5$, 5s $=5+0=5$ — all tied. Rule 2: smaller $n$ first. $3d(n=3)<4p(n=4)<5s(n=5)$ Why physically? With equal $K$, the orbital sitting in the innermost shell (smallest $n$) hugs the nucleus most tightly for its budget — the "smaller $n$" tie-break is really "the inner shell is lower energy." So 3d fills before 4p before 5s.

Removal is a different question from filling (see Ionization and Electron Removal Order). Electrons leave from the orbital of highest $n$ first. Between $4s(n=4)$ and $3d(n=3)$, the $4s$ electrons go first. Remove both $4s$ electrons: ${\text{Fe}}^{2+}=[\text{Ar}]3d^6$ Not $[\text{Ar}]3d^{4}4s^2$. Count: $[\text{Ar}]=18$, plus $3d^6=6$, gives $24=26-2$. ✅

The core is Argon ($Z=18$). Add valence electrons: $4s^2+3d^{10}+4p^3=2+10+3=15$. $Z=18+15=33(\text{Arsenic})$ The last-filled subshell is a $p$ subshell → this is a p-block element (see Periodic Table Blocks (s, p, d, f)). With $4p^3$ it is in group 15.

Actual: $\text{Cr}=[\text{Ar}]3d^{5}4s^1$ Why: A half-filled $d$ subshell ($d^5$, one electron in each of the five $d$ orbitals) is unusually stable — the electrons spread out with parallel spins (Hund's Rule), minimizing repulsion. The tiny energy gap between $4s$ and $3d$ is small enough that promoting one $4s$ electron into $3d$ to reach $d^5$ pays off. Count check: $[\text{Ar}]=18$, $+3d^5+4s^1=18+5+1=24$. ✅

Naïve Aufbau: $[\text{Ar}]3d^{9}4s^2$. Actual: $\text{Cu}=[\text{Ar}]3d^{10}4s^1$ Why: A completely filled $d^{10}$ subshell is extra stable, so one $4s$ electron drops into $3d$ to complete it. Count check: $18+10+1=29$. ✅

Five $3d$ orbitals, five electrons. By Hund's Rule, put one electron in each orbital with the same spin before any pairing begins: $\uparrow \uparrow \uparrow \uparrow \uparrow$ All five are singly occupied and parallel → 5 unpaired electrons. (Adding the lone $4s^1$ gives $6$ unpaired total for the atom, but within $3d$ it is 5.)

Forecast (Madelung order until 34 electrons): $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^4$ Verify the count: $2+2+6+2+6+2+10+4=34$. ✅ Last subshell filled is $4p$ → p-block; $4p^4$ means 4 electrons in the $p$ subshell → group 16 (a chalcogen, needs 2 more electrons for an octet). Consistent with Se being just left of Br from the parent's example.

Krypton is $Z=36$. Add valence: $5s^2+4d^{10}+5p^2=2+10+2=14$. $Z=36+14=50(\text{Tin})$ Last-filled subshell $5p$ → p-block, group 14. $+2$ ion: remove the two highest-$n$ electrons. The outermost are the $5p^2$ electrons ($n=5$), which leave first: ${\text{X}}^{2+}=[\text{Kr}]5s^{2}4d^{10}$ valence ends in $5s^2$ (the $5p$ is emptied).

• (a) $K_{6s}=6+0=6$; $K_{4f}=4+3=7$. 6s has the smaller $K$.
• (b) Smaller $K$ fills first → 6s fills before 4f.
• (c) The periodic table is literally built in Madelung order (see Periodic Table Blocks (s, p, d, f)). Because $6s(K=6)$ beats $4f(K=7)$, the $6s$ elements (Cs, Ba) are placed before the $4f$ lanthanide row. The small "4" in $4f$ is outweighed by its large $l=3$, which pushes $K$ up to 7 — the letter wins here.